Biomedical Engineering Reference
In-Depth Information
F
P
F
A
1 m
3 m
O
W
2
W
1
Unknowns:
F
A
and
F
P
. Let us choose the point O as the fulcrum. The feature that
makes this point convenient is the null lever arm of the force
F
A
and hence its
torque is zero about O. Thus, applying the second condition of equilibrium:
M
T
¼
W
2
(1.5 m)
W
1
(2.0 m) +
F
P
(3.0 m)
¼
0,
(50 kg)(10 m/s
2
)(2.0 m) +
F
P
(3.0 m),
M
T
¼
0
¼
(100 N)(1.5 m)
F
P
¼
[(150 + 1,000)N m]/3.0 m
¼
383.3 N,
F
P
¼
383.3 N.
Applying now the first condition of equilibrium that the resultant of forces must
be zero:
F
A
+
F
P
¼
W
1
+
W
2
,
F
A
¼
W
1
+
W
2
F
P
¼
500 N + 100 N
383.3 N
¼
216.7 N,
F
A
¼
216.7 N.
In Example 6.1, all of the forces that act on the board are applied on the same
segment of the straight line, that is, the board.
In Example 6.2 and Exercises 6.1 and 6.2, all of the forces that act on the body
are parallel, perpendicular to the ground, but applied on different levels in relation
to the vertical axis. To solve these exercises, all of the force vectors can be
transported to the same level, for example, to the ground, as already seen in
Chap.
1
, in vector addition.
Example 6.2
Find the magnitude of the contact forces exerted by the ground on the
right foot
N
r
and on the left foot
N
l
of a man standing erect, shown in the figure of
Example 6.2. The mass of this man is 70 kg and the location of the center of gravity
is indicated in the figure. His feet are 30 cm apart and the line that passes through
C.G. passes midway between his feet.
